1)
设al为xmol, nMg=(7.8-27x)/24 mol;
Al+3H+=Al3+ + 1.5H2;
Mg+2H+=Mg2+ + H2
nH2=1.5x+(7.8-27x)/24=8.96/22.4;x=0.2 mol nMg=0.1mol
%Al=0.2*27/(0.2*27+0.1*24)=0.6923
%Mg=1-0.6923=0.3077
2)
Al+H2O+OH-=AlO2- + 1.5H2
2.24L=0.1mol H2 => 0.1/3*2=0.067molAl
Mg+2H+=Mg2+ + H2 =>0.1molMg
%Al=0.067*27/(0.067*27+0.1*24)=0.430
%Mg=1-0.430=0.570